Let an ellipse x^2/a^2 + y^2/b^2 = 1, a < b, pass through the point (4, 3) and have eccentricity √5/3. Then the length of its latus rectum is :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q10 – JEE-MAIN 2026

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Home Maths Conic SectionsQuestion #109

Let an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

a < b

, pass through the point

(4, 3)

and have eccentricity

\frac{\sqrt{5}}{3}

. Then the length of its latus rectum is :

Clue 1 of 3: Ellipse Properties

Identify the orientation of the major axis based on the condition $a < b$ and recall the formula for eccentricity and latus rectum for such an ellipse.

🧠 Full Solution Path

Step 1: Identify Ellipse Orientation and Formulas✦ Active

The given ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with $a < b$ . This condition implies that the major axis of the ellipse is along the y-axis. For such an ellipse, the eccentricity $e$ is given by the relation $e^{2} = 1 - \frac{a^{2}}{b^{2}}$ , and the length of the latus rectum $L$ is given by $L = \frac{2 a^{2}}{b}$ .

Step 2: Use Eccentricity to Relate

a^{2}

and

b^{2}

○ Expand

We are given the eccentricity $e = \frac{\sqrt{5}}{3}$ . Squaring this, we get $e^{2} = {(\frac{\sqrt{5}}{3})}^{2} = \frac{5}{9}$ . Using the eccentricity formula for an ellipse with major axis along the y-axis:

\frac{5}{9} = 1 - \frac{a^{2}}{b^{2}}

Rearranging this equation to find a relationship between $a^{2}$ and $b^{2}$ :

\frac{a^{2}}{b^{2}} = 1 - \frac{5}{9} = \frac{4}{9} ⟹ a^{2} = \frac{4}{9} b^{2}

💡 Teacher's Secret Hint

Ensure to use the correct eccentricity formula based on the major axis orientation.

Step 3: Solve for

a^{2}

and

b^{2}

and Calculate Latus Rectum○ Expand

The ellipse passes through the point $(4, 3)$ . Substituting these coordinates into the ellipse equation:

\frac{4^{2}}{a^{2}} + \frac{3^{2}}{b^{2}} = 1 ⟹ \frac{16}{a^{2}} + \frac{9}{b^{2}} = 1

Now, substitute $a^{2} = \frac{4}{9} b^{2}$ into this equation:

\frac{16}{\frac{4}{9} b^{2}} + \frac{9}{b^{2}} = 1 ⟹ \frac{16 \times 9}{4 b^{2}} + \frac{9}{b^{2}} = 1 ⟹ \frac{36}{b^{2}} + \frac{9}{b^{2}} = 1

Combining the terms:

\frac{45}{b^{2}} = 1 ⟹ b^{2} = 45

Now find $a^{2}$ using $a^{2} = \frac{4}{9} b^{2}$ :

a^{2} = \frac{4}{9} (45) = 4 \times 5 = 20

Finally, calculate the length of the latus rectum $L = \frac{2 a^{2}}{b}$ :

L = \frac{2 (20)}{\sqrt{45}} = \frac{40}{3 \sqrt{5}}

Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{5}$ :

L = \frac{40 \sqrt{5}}{3 \sqrt{5} \times \sqrt{5}} = \frac{40 \sqrt{5}}{3 \times 5} = \frac{40 \sqrt{5}}{15} = \frac{8 \sqrt{5}}{3}

💡 Teacher's Secret Hint

Remember to rationalize the denominator for the final answer.

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Maths Question 10 – JEE-MAIN 2026

Clue 1 of 3: Ellipse Properties

🧠 Full Solution Path